Question

1.) Resolva o sistema de tres equações aplicando a Regra de cramer


2x-y+z=2

x+y-z=0

3x-2y+3z=4

Answer 1

Para resolver o sistema utilizando a regra de Cramer, primeiro liste todos os determinantes necessários

$\begin{gathered}\begin{cases} {2x - y + z = 2}\\{x + y - z = 0} \\ 3x - 2y + 3z = 4\end{cases}\end{gathered} \\ D=\begin{gathered}\left|\begin{array}{ccc}2&-1&1\\1&1& - 1\\3&-2&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} \\ D_{1} = \begin{gathered}\left|\begin{array}{ccc}2&-1&1\\0&1& - 1\\4&-2&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} \\ D_{2} = \begin{gathered}\left|\begin{array}{ccc}2&2&1\\1&0& - 1\\3&4&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} \\ D_{3} = \begin{gathered}\left|\begin{array}{ccc}2&-1&2\\1&1&0\\3& - 2&4\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered}$

$D=\begin{gathered}\left|\begin{array}{ccc}2&-1&1\\1&1& - 1\\3&-2&3\end{array}\right|~\begin{matrix}2& - 1 \\ 1&1 \\ 3 &- 2\end{matrix}\end{gathered}$

Usando a regra de Sarrus, some os produtos das diagonais que vão de cima para baixo e subtraia os produtos das diagonais que vão de baixo para cima

$2 \times 1 \times 3 + ( - 1) \times ( - 1) \times 3 + 1 \times 1 \times ( - 2) - (3 \times 1 \times 1 + ( - 2) \times ( - 1) \times 2 + 3 \times 1 \times ( - 1)) \\ D =3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$D_{1} = - 0 \times \begin{gathered}\left|\begin{array}{ccc} - 1&1\\ - 2&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} + 1 \times \begin{gathered}\left|\begin{array}{ccc}2&1\\4&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} - ( - 1) \times \begin{gathered}\left|\begin{array}{ccc}2&-1\\4&-2\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} \\ D_{1} = 2$

$D_{2} = - 1 \times \begin{gathered}\left|\begin{array}{ccc}2&1\\4&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} + 0 \times \begin{gathered}\left|\begin{array}{ccc}2&1\\3&3\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} - ( - 1) \times \begin{gathered}\left|\begin{array}{ccc}2&2\\3&4\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} \\ D_{2} = 0$

$D_{3} = - 1 \times \begin{gathered}\left|\begin{array}{ccc} - 1&2\\ - 2&4\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} + 1 \times \begin{gathered}\left|\begin{array}{ccc}2&2\\3&4\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} - 0 \times \begin{gathered}\left|\begin{array}{ccc}2&-1\\3&-2\end{array}\right|~\begin{matrix}\end{matrix}\end{gathered} \\ D_{3} = 2$

Dado D ≠ 0, a regra Cramer pode ser aplicada, então encontre x, y, z utilizando as fórmulas $x = \frac{D_{1}}{D} \\ \\ y = \frac{D_{2}}{D} \\ \\ z = \frac{D_{3}}{D}$

$x = \frac{2}{3} \\ \\ y = \frac{0}{3} = 0 \\ \\ z = \frac{2}{3}$

$S=\left({ x,y,z }\right) = \left({  \frac{2}{3}, 0 ,\frac{2}{3} }\right)$

$\mathcal{Bons \: estudos } \\ \displaystyle\int_ \empty ^ \mathbb{C}     \frac{ - b \: ± \:  \sqrt{ {b}^{2} - 4 \times a \times c } }{2 \times a} d{ t } \boxed{ \boxed{ \mathbb{\displaystyle\Re}\sf{ \gamma  \alpha }\tt{ \pi}\bf{ \nabla}}}$